Find all real values of $x$ which satisfy
\[\frac{1}{x + 1} + \frac{6}{x + 5} \ge 1.\]
Solution: Subtracting 1 from both sides and putting everything over a common denominator, we get
\[\frac{-x^2 + x + 6}{(x + 1)(x + 5)} \ge 0.\]Equivalently,
\[\frac{x^2 - x - 6}{(x + 1)(x + 5)} \le 0.\]We can factor the numerator, to get
\[\frac{(x - 3)(x + 2)}{(x + 1)(x + 5)} \le 0.\]We build a sign chart, accordingly.
\begin{tabular}{c|cccc|c} &$x-3$ &$x+2$ &$x+1$ &$x+5$ &$f(x)$ \\ \hline$x<-5$ &$-$&$-$&$-$&$-$&$+$\\ [.1cm]$-5<x<-2$ &$-$&$-$&$-$&$+$&$-$\\ [.1cm]$-2<x<-1$ &$-$&$+$&$-$&$+$&$+$\\ [.1cm]$-1<x<3$ &$-$&$+$&$+$&$+$&$-$\\ [.1cm]$x>3$ &$+$&$+$&$+$&$+$&$+$\\ [.1cm]\end{tabular}Also, note that $\frac{(x - 3)(x + 2)}{(x + 1)(x + 5)} = 0$ for $x = -2$ and $x = 3.$  Therefore, the solution is
\[x \in \boxed{(-5,-2] \cup (-1,3]}.\]